Let f(x) be a periodic function of x with period 3 and f(x) = |2x − 1| + x for


−1 < x ≤ 2. Sketch the graph of the curve y = f(x) in the interval [−3, 4]

To graph \( y = f(x) \), we will follow these steps:

### Step 1: Understanding the function and periodicity
- The function \( f(x) \) is periodic with a period of 3, which means that the graph of \( f(x) \) repeats every 3 units.
- For the interval \( -1 < x \leq 2 \), the function is defined as:
  \[
  f(x) = |2x - 1| + x
  \]
  
We will first sketch this function on \( -1 < x \leq 2 \), and then use periodicity to repeat the pattern over the interval \( [-3, 4] \).

### Step 2: Analyzing \( f(x) = |2x - 1| + x \)
- \( |2x - 1| \) changes behavior at \( x = \frac{1}{2} \), since that’s where the absolute value function shifts from negative to positive.
- So, we can break down \( f(x) \) into two pieces:
  - For \( x \leq \frac{1}{2} \), \( |2x - 1| = -(2x - 1) = -2x + 1 \).
    \[
    f(x) = (-2x + 1) + x = -x + 1
    \]
  - For \( x > \frac{1}{2} \), \( |2x - 1| = 2x - 1 \).
    \[
    f(x) = (2x - 1) + x = 3x - 1
    \]

Thus, \( f(x) \) is defined piecewise as:
\[
f(x) =
\begin{cases}
-x + 1 & \text{for } -1 < x \leq \frac{1}{2}, \\
3x - 1 & \text{for } \frac{1}{2} < x \leq 2.
\end{cases}
\]

### Step 3: Plotting the function for \( -1 < x \leq 2 \)
Now, let’s calculate some points for \( f(x) \) in the interval \( -1 < x \leq 2 \):

- For \( x = -1 \), \( f(-1) = (-(-1) + 1) = 2 \).
- For \( x = 0 \), \( f(0) = (-0 + 1) = 1 \).
- For \( x = \frac{1}{2} \), \( f\left(\frac{1}{2}\right) = -\frac{1}{2} + 1 = \frac{1}{2} \).
- For \( x = 1 \), \( f(1) = (3(1) - 1) = 2 \).
- For \( x = 2 \), \( f(2) = (3(2) - 1) = 5 \).

We now have the following points:
- \( (-1, 2), (0, 1), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 2), (2, 5) \).

We can plot these points and connect them using straight lines.

### Step 4: Extending the graph using periodicity
Since \( f(x) \) is periodic with a period of 3, we repeat the pattern from \( -1 < x \leq 2 \) over the entire interval \( [-3, 4] \). 

- Repeat the pattern for the interval \( [-4, -1] \) and \( [2, 5] \), ensuring that the graph is continuous where the pieces meet.

#### Final Graph:
The graph will consist of piecewise linear segments based on the pattern derived above, repeating every 3 units.

Would you like me to generate a sketch of this graph?

### Follow-Up Questions
1. What would happen if the period of \( f(x) \) were changed to 2 instead of 3?
2. How would the graph change if we altered the function to \( f(x) = |3x - 2| + x \)?
3. Can you calculate the values of \( f(x) \) at \( x = 3 \) and \( x = -2 \)?
4. How would adding a constant \( c \) to the function (i.e., \( f(x) + c \)) affect the graph?
5. Can you describe the effect of reflecting the graph of \( f(x) \) over the x-axis?

### Tip:
When dealing with periodic functions, it's helpful to fully analyze the behavior of the function over one period first, then replicate that pattern across the desired interval.

Let f(x) be a periodic function of x with period 3 and f(x) = |2x − 1| + x for −1 < x ≤ 2. Sketch the graph of the curve y = f(x) in the interval [−3, 4].

Explore how to graph the periodic function f(x) = |2x - 1| + x with a period of 3 over the interval [-3, 4]. This solution covers key concepts in periodicity, absolute value, and piecewise functions, suitable for students in Grades 10-12.

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